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Proof step 3 from proof diff-empty-subseteq (previous step)

⊦ (∀ adr. adr ∈ as0 \ as1 ↔ adr ∈ ∅) ↔ as0 ⊆ as1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∀ adr. ¬ adr ∈ as0 \ as1) ↔ as0 ⊆ as1

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ ∅; (s)